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Question

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(a) If x is the your distance from the wall holding the painting, find a function A(x) that computes the angle a in the picture; A(x)=
(b) Find the rate of change with respect to x of the angle a when you are at the location x
(c) Find the rate of change with respect to x of the angle a when you are at the location x=4 m. (Exact answer.)

(d) Find the rate of change with respect to x of the angle a when you are at the location x=13 m. (Exact answer.)

(e) Find the location where the rate of change with respect to x of the angle a is zero m.(Exact answer.)
(f) When you are standing at the location to maximize a, what is the value of a? (Radians to three decimal places.)

Explanation / Answer

A)

A(x) = arctan(13/x) - arctan(3/x)

B)

A'(x) = -(10 (-39+x^2))/((9+x^2) (169+x^2))

C)

-(10 (-39+4^2))/((9+4^2) (169+4^2)) = 46/925

D)

-(10 (-39+13^2))/((9+13^2) (169+13^2)) = -25/1157

E)

Solve: -(10 (-39+x^2))/((9+x^2) (169+x^2)) = 0

x = sqrt(39)

F)

arctan(13/sqrt(39)) - arctan(3/sqrt(39)) =arctan(5/sqrt(39)) = 0.67513153

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